help velocity and mass given


by Vick007
Tags: equations, friction, horizontal, mass, velocity
Vick007
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#1
Dec18-11, 08:48 PM
P: 9
1. The problem statement, all variables and given/known data

With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )

2. Relevant equations
I don't know how to use the mass in any equations. help.


3. The attempt at a solution
something with xo= vo + 1/2at^2 ?? how is the mass used?
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gneill
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#2
Dec18-11, 08:53 PM
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Quote Quote by Vick007 View Post
1. The problem statement, all variables and given/known data

With an initial speed of 5.00 m/s a 1.00 kg disk is slid on a horizontal steel plate. How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )

2. Relevant equations
I don't know how to use the mass in any equations. help.


3. The attempt at a solution
something with xo= vo + 1/2at^2 ?? how is the mass used?
The question appears to be incomplete. You need to know the coefficient of dynamic friction between the disk and steel plate in order to proceed. You'll use the coefficient of friction and the mass to determine the force that retards the motion, hence the acceleration.
Vick007
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#3
Dec19-11, 12:42 AM
P: 9
ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks

gneill
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#4
Dec19-11, 07:44 AM
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help velocity and mass given


Quote Quote by Vick007 View Post
ok so I checked and yeah the coefficient is given as 0.04 what do I do now? thanks
What is the relationship between the coefficient of friction and the frictional force?
Vick007
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#5
Dec19-11, 11:41 AM
P: 9
weight X coefficient of friction = friction force?
how does that tie in with the velocity given?
gneill
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#6
Dec19-11, 12:36 PM
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Quote Quote by Vick007 View Post
weight X coefficient of friction = friction force?
Yes.
how does that tie in with the velocity given?
Draw the free body diagram for the disk. What forces are acting? What is the acceleration of the disk (hint: Newton's second law).
Vick007
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#7
Dec20-11, 09:03 AM
P: 9
Ff= force of friction
U= coefficient of friction
Fn= force normal
m=mass
g=acceleration due to gravity =9.81
a= acceleration
d=distance

Ff=UxFn
which is the same as
m x a= U x m x -g
divide out the m and you get
a=U x -g
and so
a=.04 x -9.81
a= -.3924
so is it slowing down by .3924 m/s^2?
how do I link the velocity, time, and acceleration now?
gneill
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#8
Dec20-11, 09:31 AM
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Quote Quote by Vick007 View Post
Ff= force of friction
U= coefficient of friction
Fn= force normal
m=mass
g=acceleration due to gravity =9.81
a= acceleration
d=distance

Ff=UxFn
which is the same as
m x a= U x m x -g
divide out the m and you get
a=U x -g
and so
a=.04 x -9.81
a= -.3924
so is it slowing down by .3924 m/s^2?
how do I link the velocity, time, and acceleration now?
What's the relationship between acceleration and velocity? How about if there's an initial velocity?
Vick007
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#9
Dec20-11, 09:37 AM
P: 9
the initial velocity is 5.00 m/s and since the acceleration is (I think) -.3924 and I'm looking for the time would this equation work?
a = (V2 - V1)/T
where
(V2 - V1) = change in velocity
so it would say
-.3924 = (V2 - 5.00)/T
but what is V2? or how do I find it?
gneill
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#10
Dec20-11, 10:10 AM
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Quote Quote by Vick007 View Post
the initial velocity is 5.00 m/s and since the acceleration is (I think) -.3924 and I'm looking for the time would this equation work?
a = (V2 - V1)/T
That will work. It's a rearrangement of the usual equation, v2 = v1 + a*t .
where
(V2 - V1) = change in velocity
so it would say
-.3924 = (V2 - 5.00)/T
but what is V2? or how do I find it?
What is the speed of the puck when it stops?
Vick007
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#11
Dec20-11, 11:04 AM
P: 9
so V2 = 0
and now I just have to find T
so according to your equation it would look like this:
v2 = v1 + a*t
and if I fill it in I get
0 = 5.00 + (-.3924) * t
-5 = -.3924 * t
(-5/-.3924) = t
t = 12.42 sec (is that correct?)
thanks a lot for helping, but there's a second part to the problem
How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )
how do I begin the second part now that I have the time? Is there some equation?
gneill
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#12
Dec20-11, 01:20 PM
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Quote Quote by Vick007 View Post
so V2 = 0
and now I just have to find T
so according to your equation it would look like this:
v2 = v1 + a*t
and if I fill it in I get
0 = 5.00 + (-.3924) * t
-5 = -.3924 * t
(-5/-.3924) = t
t = 12.42 sec (is that correct?)
Probably a typo, t = 12.74 should be your value.
thanks a lot for helping, but there's a second part to the problem
How long will it take for the disk to stop, and how far does it travel? ( do not neglect friction )
how do I begin the second part now that I have the time? Is there some equation?
I'm rather surprised that you don't seem to have had any introduction to the basic kinematic formulas. Yes there is an equation (there's always an equation :smile ). In this case you want one that relates distance traveled to initial velocity, acceleration, and time. Any ideas as to which of the standard kinematic equations might fit?
Vick007
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#13
Dec20-11, 02:15 PM
P: 9
would it be one like this?
X - Xo = .5(Vo + V)t
which could be simplified to
X = .5(5+V)(12.74)
how do I find the velocity though?
gneill
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#14
Dec20-11, 02:30 PM
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Quote Quote by Vick007 View Post
would it be one like this?
X - Xo = .5(Vo + V)t
which could be simplified to
X = .5(5+V)(12.74)
how do I find the velocity though?
That equation will work. It incorporates the initial velocity (Vo) and the final velocity (V = 0) in order to take care of the acceleration. Remember that you previously wrote: a = (V2 - V1)/T.

So plug in your V=0 and find distance X.

Another equation that you should become familiar with is:

[itex]d = v_o t + \frac{1}{2} a t^2 [/itex]

See the Physics Formulary
Vick007
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#15
Dec20-11, 06:47 PM
P: 9
ok so I got like
X = 31.86 meters?
is that correct?
gneill
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#16
Dec20-11, 09:40 PM
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Quote Quote by Vick007 View Post
ok so I got like
X = 31.86 meters?
is that correct?
Yes, that is correct.


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